For the capacitor network shown in the Figure , the potential difference across ab is 220 V.
A)Find the total charge stored in this network
B) Find the charge on each capacitor.
Enter your answer as two numbers, separated with a comma.
C) Find the total energy stored in the network.
D) Find the energy stored in each capacitor.
Enter your answers numerically separated by a comma.
E) Find the potential difference across each capacitor.
Enter your answers numerically separated by a comma.
2 Answers

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As 35 nF and 75 nF are in parallel their equivalent capacity ‘C’ is 110 nF
A)The total charge stored in this network=Q=CV=110*220*10^9
The total charge stored in this network=Q=CV=2.42*10^5 C
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B) The charge on 35 nF =7.7*10^6 C.
The charge on 75 nF =1.65*10^ 5 C.
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C) The total energy stored U =(1/2)CV^2=2.662*10^3 J
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D) The energy stored in 35 nF capacitor.=8.47*10^4 J
The energy stored in 75 nF capacitor.=1.815*10^3 J
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E) The potential difference across each capacitor is 220 V
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I always remember these formulas which are for DC based on the resistor first. In series Rt = R1+R2+R3… In parallel 1/Rt = 1/R1+1/R2+1/R3 Since resistors and coils (inductors) have effect but pass electricity the formulas work the same. However capacitors block electricity, so the same formulas in reversed situations.